It has no solution
18m+7=18m+7
Which means they are the same equation !!
Answer:
See the step by step solution
Step-by-step explanation:
Consider a more general case of the form 
. We want to express it as a sine term only by using the following formula 
The trick consists on find a number ![\beta \in [0,2\pi]](https://tex.z-dn.net/?f=%5Cbeta%20%5Cin%20%5B0%2C2%5Cpi%5D)
such that 
. But, note that since 
it is most likely that 
. Then, we will use the following equations:
![\cos(\beta) =\frac{a}{\sqrt[]{a^2+b^2}=A,\sin(\beta) =\frac{b}{\sqrt[]{a^2+b^2}=B](https://tex.z-dn.net/?f=%5Ccos%28%5Cbeta%29%20%3D%5Cfrac%7Ba%7D%7B%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%3DA%2C%5Csin%28%5Cbeta%29%20%3D%5Cfrac%7Bb%7D%7B%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%3DB)
Then, problem turns out to be
![a\sin(x)+b\cos(x)= \sqrt[]{a^2+b^2}(A\sin(x)+B\cos(x)) =\sqrt[]{a^2+b^2}(\sin(x)\cos(\beta)+\sin(\beta)\cos(x))=\sqrt[]{a^2+b^2}\sin(x+\beta)](https://tex.z-dn.net/?f=a%5Csin%28x%29%2Bb%5Ccos%28x%29%3D%20%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%28A%5Csin%28x%29%2BB%5Ccos%28x%29%29%20%3D%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%28%5Csin%28x%29%5Ccos%28%5Cbeta%29%2B%5Csin%28%5Cbeta%29%5Ccos%28x%29%29%3D%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%5Csin%28x%2B%5Cbeta%29%20)
,
where ![\beta = \tan^{-1}(\frac{B}{A})= \tan^{-1}(\frac{b}{a}). So, note that the frequency is the same. Then, a) Taking a=4 and b= 15 we have that 4 sin 2πt + 15 cos 2πt= [tex]\sqrt[]{15^2+4^2}\sin(2\pi t + \tan^{-1}\frac{15}{4})](https://tex.z-dn.net/?f=%5Cbeta%20%3D%20%5Ctan%5E%7B-1%7D%28%5Cfrac%7BB%7D%7BA%7D%29%3D%20%5Ctan%5E%7B-1%7D%28%5Cfrac%7Bb%7D%7Ba%7D%29.%20So%2C%20note%20that%20the%20frequency%20is%20the%20same.%20%3C%2Fp%3E%3Cp%3EThen%2C%20%3C%2Fp%3E%3Cp%3Ea%29%20Taking%20a%3D4%20and%20b%3D%2015%20we%20have%20that%204%20sin%202%CF%80t%20%2B%2015%20cos%202%CF%80t%3D%20%5Btex%5D%5Csqrt%5B%5D%7B15%5E2%2B4%5E2%7D%5Csin%282%5Cpi%20t%20%2B%20%5Ctan%5E%7B-1%7D%5Cfrac%7B15%7D%7B4%7D%29)
b) The amplitude is ![\sqrt[]{15^2+4^2} = \sqrt[]{241}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B15%5E2%2B4%5E2%7D%20%3D%20%5Csqrt%5B%5D%7B241%7D)
. Since 
the frequency is 1 Hz. The period is given by \frac{2\pi}{2\pi} = 1.
c) The function's graph is attached
Complete Question:
Convert 3x-y+2=0 in rectangle form to polar form
Answer:
r ( 3cosθ - sinθ) = -2
Step-by-step explanation:
The equation so given is already in rectangular form, the task is to convert it to polar form.
3x - y + 2 = 0
To transform an equation from the rectangular to the polar coordinate:
x = r cosθ
y = r sinθ
Substitute the representations for x and y into the given equation:
3(r cosθ) - (r sinθ) + 2 = 0
3rcosθ - rsinθ = -2
The polar representation of the equation then becomes:
r ( 3cosθ - sinθ) = -2
Answer:
22x + 24
Step-by-step explanation:
7x+9+7x+9+4x+3+4x+3
11x + 18 + 11x + 6
Mark brainliest please!
128 square inches.
The area of a triangle is 1/2bh, where b is the base and h is the height. Substitute the values in.
1/2 * 8 * 32
Now, just solve.
4 * 32
128 square inches
Please consider marking this answer as Brainliest to help me advance.
